Question

`Q.1` Find the values of K so that the quadratic equation`x^2+2(K+1)x+K+5=0` has at least one positive root.?

Answer 1

`K=-5`

Explanation:

A quadratic equation basic formula is
`ax^2+bx+c`
so matching up with `x^2+2(K+1)x+k+5=0`
gives
`x^2` terms: `x^2` `:.a=1`
`x` terms: `2(K+1)` or `2K+2` `:.b=2K+2`
constants: `K+5 :. c=K+5`

For a quadratic to have at least one positive root, the discriminate `(Delta)` must be equal to (1 real root) or greater than (two real roots) 0.

From the quadratic formula, the discriminate is `b^2-4ac`. We need `b^2≥4ac`

So in this case, `(2K+2)^2≥(4*1*(K+5))`
`:.4K^2+4K+4≥4K+20`
`K^2+K+1≥K+5`
`K^2+1≥+5`
`K≥4`

So if K=4 then
`x^2+2(4+1)x+(5+4)=0`
`x^2+10x+9=0`
`(x+9)(x+1) :. x=-9 and x=-1` These are both negative so `K!=4`

It seems like we need the c part to cancel. So if `K=-5` then
the equation would be
`x^2+2(-5+1)x+(5-5)=0`
`x^2-8x+0=0`
`x(x-8)`
`:. x=0 and x=8`

8 is +ve so K=-5 must be the solution.